Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 23 - Gauss' Law - Problems - Page 684: 76b

Answer

$E = \frac{\rho~R^2}{2~\epsilon_0~r}$

Work Step by Step

We can draw a Gaussian cylinder of length $L$ with the axis along the same axis as the charged cylinder. We can find the electric field at a distance $r$ from the cylinder axis where $r \gt R$: $\epsilon_0~\Phi = q_{enc}$ $(\epsilon_0)~(E)~(2\pi~r~L) = \rho~\pi~R^2~L$ $(\epsilon_0)~(E)~(2~r) = \rho~R^2$ $E = \frac{\rho~R^2}{2~\epsilon_0~r}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.